A current `I` flows along a round circular loop of radius `R`. Find the line integration of magnetic field along the axis of the loop from center to `oo`

Options

A. `mu_(0)I`
B. `4mu_(0)I`
C. `(mu_(0)I)/2`
D. `(mu_(0)I)/4`

Explanation

To find the line integration of the magnetic field along the axis of a circular loop from the center to infinity, we can follow these steps:

### Step 1: Determine the Magnetic Field at a Distance x from the Center
The magnetic field \( B \) at a distance \( x \) along the axis of a circular loop of radius \( R \) carrying a current \( I \) is given by the formula:

\[ B(x) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

where \( \mu_0 \) is the permeability of free space.

### Step 2: Set Up the Line Integral
We want to find the line integral of the magnetic field from the center of the loop (where \( x = 0 \)) to infinity (where \( x \to \infty \)). The line integral \( L \) can be expressed as:

\[ L = \int_{0}^{\infty} B(x) \, dx \]

Substituting the expression for \( B(x) \):

\[ L = \int_{0}^{\infty} \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \, dx \]

### Step 3: Simplify the Integral
Since \( \mu_0 \), \( I \), and \( R^2 \) are constants, we can factor them out of the integral:

\[ L = \frac{\mu_0 I R^2}{2} \int_{0}^{\infty} \frac{1}{(R^2 + x^2)^{3/2}} \, dx \]

### Step 4: Use Substitution for the Integral
To evaluate the integral, we can use the substitution \( x = R \tan(\theta) \). Then, \( dx = R \sec^2(\theta) \, d\theta \) and the limits change as follows:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x \to \infty \), \( \theta \to \frac{\pi}{2} \)

Substituting these into the integral gives:

\[ L = \frac{\mu_0 I R^2}{2} \int_{0}^{\frac{\pi}{2}} \frac{R \sec^2(\theta)}{(R^2 + R^2 \tan^2(\theta))^{3/2}} \, d\theta \]

This simplifies to:

\[ L = \frac{\mu_0 I R^2}{2} \int_{0}^{\frac{\pi}{2}} \frac{R \sec^2(\theta)}{(R^2 \sec^2(\theta))^{3/2}} \, d\theta \]

### Step 5: Simplify Further
This can be simplified to:

\[ L = \frac{\mu_0 I R^2}{2} \int_{0}^{\frac{\pi}{2}} \frac{R \sec^2(\theta)}{R^3 \sec^3(\theta)} \, d\theta = \frac{\mu_0 I}{2R} \int_{0}^{\frac{\pi}{2}} \cos(\theta) \, d\theta \]

### Step 6: Evaluate the Integral
The integral \( \int_{0}^{\frac{\pi}{2}} \cos(\theta) \, d\theta \) evaluates to 1:

\[ L = \frac{\mu_0 I}{2R} \cdot 1 = \frac{\mu_0 I}{2R} \]

### Final Answer
Thus, the line integral of the magnetic field along the axis of the loop from the center to infinity is:

\[ \boxed{\frac{\mu_0 I}{2R}} \]